PUNJAB PUBLIC SERVICE COMMISSION WRITTEN TEST FOR THE POST OF LECTURER IN MATHEMATICS 2015

Differential Equation Related Material

Q.01 $\fn_cm \int_{-4}^{0}\frac{t}{\sqrt{16-t^2}}dt=$ __________________

A. 0

B. Divergent

C. -4

D. 4

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$\fn_cm \int_{-4}^{0}\frac{t}{\sqrt{16-t^2}}dt=-\frac{1}{2}\int_{-4}^{0}\frac{-2t}{\sqrt{16-t^2}}dt=-\sqrt{16-t^2}|_{-4}^{0}=-4$

Q.02 The period of function $\fn_cm Acos\omega t+Bsin\omega t$ is

A. $\fn_cm \frac{\omega}{2\pi}$

B. $\fn_cm 2\pi \omega$

C. $\fn_cm \frac{2\omega}{\pi}$

D. $\fn_cm \frac{2\pi}{\omega}$

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As sint and cost are periodic function with period 2π
So
Sinwt=sin(wt+2π)=sin(t+2π/w). Similarly for cost .
Hence period of given function if 2π/w
===> D is correct

Q.03 $\inline \fn_cm A= (-4x-3y+az)i+(bx+3y+5z)j+(4x+cy+3z)k$ is irrotational if a,b,c are__________

A. 4,-3,5

B. 4,5,-3

C. -3,4,5

D. 2,3,5

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Flow is irrational if $\fn_cm \nabla \times A=0\\ \\ \begin{bmatrix} i& j & k\\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ (-4x-3y+az)& (bx+3y+5z) & (4x+cy+3z) \end{bmatrix}=0\\ \\(c-5)i-(4-a)j+(b+3)k=0i+0j+0k\\ \\ (a,b,c)=(4,-3,5)$

Q.04 $\fn_cm V= (-4x-6y+3z)i+(-2x+y-5z)j+(5x+6y+az)k$ is solenoidal if a = ________

A. 1

B. 2

C. 3

4. 4

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For solenoidal vector $\fn_cm \nabla$ .V=0
After applying partial derivative we get
-4+1+a=0
So a=3
C is correct

Q.05 $\fn_cm \fn_cm \int_{(0,0)}^{(2,1)} (10x^4-2xy^3)dx-3x^2y^2$ along the path $\fn_cm x^4-6xy^3=4y^2 is$

A. 56

B. 60

C. 62

D. 64

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We know that

$\fn_cm M=(10x^4-2xy^3),N=-3x^2y^2\\ \\ \\ \frac{\partial M}{\partial y}=-6x^2\\\\\\ \frac{\partial N}{\partial x}=-6x^2$

As the solution is exact. So we have

$\fn_cm \int_{(0,0)}^{(2,1)} (10x^4-2xy^3)dx-independent \; terms \; of \; x\\\\\\ \int_{(0,0)}^{(2,1)} (10x^4-2xy^3)dx-0 \\\\\\ ={2x^5}-x^2y^3|_{(0,0)}^{(2,1)}\\\\\\ ={64}-{4}=60$

Q.06 If S is the closed surface and v is the volume enclosed by S then $\fn_cm \int \int_s r.nds=$
A. v
B. 2v
C. 3v
D. 4v

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By Gauss Divergence Theorem

$\fn_cm \int \int_s F.\widehat{N}ds=\int \int_r \int dN.Fdv$

As position vector $\fn_cm \overrightarrow{r}= x i+yj+zk$

So $\fn_cm \int \int_s r.nds=\int \int_v \int \nabla.\overrightarrow{r}dv\\\\\\ =\int \int_v \int (\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k).(xi+yj+zk)dv\\\\\\ =\int \int_v \int(1+1+1)dv\\\\\\ =3\int \int_v \int dv =3v$

Q.07 Centrifugal acceleration is

A. $\fn_cm -\omega \times (\omega \times r)$

B. $\fn_cm \omega \times (\omega \times r)$
C. $\fn_cm \omega . (\omega \times r)$
D. $\fn_cm r \times (\omega \times r)$

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As $\fn_cm \overrightarrow{r}= cos \omega t + sin\omega t\\\\\\ \frac {d \overrightarrow{r}}{dt}=\overrightarrow{v}=- \omega sin \omega t +\omega cos\omega t\\\\\\ \frac {d \overrightarrow{v}}{dt}=\overrightarrow{a}=- \omega^2 cos \omega t -\omega^2 sin\omega t\\\\\\ \overrightarrow{a}=-\omega^2 \overrightarrow{r}=-\omega \times( \omega \times \overrightarrow{r} )$